Q.A copper wire of cross-sectional area 0.5 mm$^2$ carries a current of 0.2 A. If the free electron density of copper wire is $8.4 \times 10^{28}$ m$^{-3}$, then compute the drift velocity of free electron.

Showing the 12 most recent of 31 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.Drift velocity Vd varies with the intensity of electric field E as per the relation

   (a) Vd is proportional to E^2

   (b) Vd is proportional to 1/E

   (c) Vd is proportional to sqrt(E)

   (d) Vd is proportional to E

   ›Reveal solutionSolution

   Drift velocity is the (small) average velocity electrons gain between collisions due to the electric field, and it comes out directly proportional to E.

   When an electric field E is applied to a conductor, each free electron experiences a force F = eE, giving it an acceleration a = eE/m between collisions with the lattice ions. If tau is the average time between collisions (relaxation time), the average extra velocity gained (the drift velocity) is

   vd = a * tau = (eE/m) * tau = (e*tau/m) * E

   …

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2. CBSE 2026Set ANNUAL1 mark
   Q.If the current flowing in a copper wire be allowed to flow in another copper wire of same length but of doubled the radius then what will be the effect on the drift velocity of the electron?

   ›Reveal solutionSolution

   For the same current, $v_d \propto 1/A$, and doubling the radius quadruples the cross-sectional area.

   Current is related to drift velocity by $I = nAev_d$, so for the same current $I$ (and the same material, hence the same $n$), $v_d = \dfrac{I}{nAe} \propto \dfrac{1}{A}$. If the radius is doubled, the cross-sectional area $A=\pi r^2$ becomes 4 times larger. So the drift velocity becomes

   …

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3. CBSE 2026Set ANNUAL1 mark
   Q.State Ohm's law in terms of current density, specific conductance and electric field intensity.

   ›Reveal solutionSolution

   Microscopic Ohm's law: current density J = σE (σ = conductivity, E = field).

   The usual Ohm's law is V = IR. In microscopic (vector) form, it relates the current density J (current per unit cross-sectional area) to the electric field E inside the conductor through the material's specific conductance (conductivity) σ:

   J = σ E.

   …

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4. CBSE 2026Set SEM31 markMCQ
   Q.Which of the following statement(s) is/are true ? A potential difference of V is applied at the two ends of a conductor of length l and area of cross-section A. Statement I : When potential difference is doubled, current density also gets doubled. Statement II : When potential difference is doubled, drift velocity gets halved. Statement III : When area of cross-section is doubled, current density decreases.

   (a) I and II are true

   (b) Only I is true

   (c) Only III is true

   (d) II and III are true

   ›Reveal solutionSolution

   Doubling V doubles E, so J = σE and v_d = μE both double — Statement I true, Statement II (drift velocity halved) false. J = V/(ρl) is independent of area, so Statement III (J decreases when A doubles) is also false. Only I is true → option (b).

   Statement I: J = σE and E = V/l, so doubling V doubles E and hence doubles the current density J. TRUE.

   Statement II: drift velocity v_d = (eE/m)τ ∝ E ∝ V. Doubling V doubles v_d, it does not halve it. FALSE.

   …

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5. CBSE 2025Set D1 markMCQ
   Q.The relation between drift velocity v of free electrons in conductor in electric conduction and potential difference V between ends of conductor is (A) proportional to V (B) inversely proportional to V (C) proportional to V^2 (D) inversely proportional to V^2

   ›Reveal solutionSolution

   Drift velocity is directly proportional to the potential difference V.

   In a conductor of length L across which a potential difference V is applied, the electric field is E = V/L. Free electrons acquire a drift velocity

   $v_d = \frac{eE\tau}{m} = \frac{eV\tau}{mL}$ …

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6. CBSE 2025Set D1 markMCQ
   Q.If the length of a conductor is doubled while keeping the potential difference across it constant, then the drift velocity of electron will (A) remain the same (B) be double (C) be halved (D) increase fourfold

   ›Reveal solutionSolution

   Doubling the length at constant V halves the drift velocity.

   The drift velocity is

   $v_d = \frac{eE\tau}{m} = \frac{eV\tau}{mL}$ …

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7. CBSE 2025Set ANNUAL1 markMCQ
   Q.A thick wire is stretched so that its length becomes two times. What is the ratio of change in resistance of the wire to the initial resistance of the wire?

   (i) 2 : 1

   (ii) 4 : 1

   (iii) 3 : 1

   (iv) 1 : 4

   ›Reveal solutionSolution

   New resistance is 4 times the old, so the change is 3 times the original: ratio 3 : 1.

   Resistance $R = \rho L/A$. Stretching keeps the volume $AL$ constant, so if length doubles ($L \to 2L$) the area halves ($A \to A/2$). Then $R' = \rho (2L)/(A/2) = 4\rho L/A = 4R$. The change in …

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8. CBSE 2025Set ANNUAL1 markMCQ
   Q.Calculate the amount of charge flowing in 2 minutes in a wire of resistance 10 ohm when a potential difference of 20 volts is applied between its ends.

   (i) 120 C

   (ii) 240 C

   (iii) 20 C

   (iv) 4 C

   ›Reveal solutionSolution

   Q = It = (V/R) x t = 2 A x 120 s = 240 C.

   …

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9. CBSE 2024Set 55/2/11 markMCQ
   Q.Electrons drift with speed $v_d$ in a conductor with potential difference $V$ across its ends. If $V$ is reduced to $\dfrac{V}{2}$, their drift speed will become : (A) $\dfrac{v_d}{2}$ (B) $v_d$ (C) $2v_d$ (D) $4v_d$

   ›Reveal solutionSolution

   Drift speed is directly proportional to the applied potential difference for a given conductor, so halving $V$ halves $v_d$. The new drift speed is $\dfrac{v_d}{2}$, which is option (A).

   The key here is understanding what drift speed actually depends on. Many students memorise the formula $v_d = \frac{I}{n e A}$ and then try to relate $I$ to $V$ via Ohm's law — that works, but it's easy to lose track of which quantities stay constant. Let's build it from the physics up.

   Drift speed is the average velocity electrons acquire due to an electric field inside the conductor. That field is $E = V/L$, where $L$ is the length of the conductor. The force on each electron is $eE$, and in the steady state, this force is balanced by collisions with the lattice, giving a constant drift speed proportional to the field. So the fundamental proportionality is:

   $v_d \propto E \quad \text{and since} \quad E = \frac{V}{L}, \quad \text{we get} \quad v_d \propto V$

   for a fixed conductor (fixed $L$, fixed material properties like relaxation time $\tau$, mass $m$, charge $e$).

   Now let's walk through it step by step.

   1. Start with the microscopic relation. The drift speed is given by $v_d = \frac{eE\tau}{m}$, where $\tau$ is the average time between collisions (relaxation time). This comes from $F = eE = m a$, and then $v_d = a\tau$. For a given conductor at a fixed temperature, $\tau$, $m$, and $e$ are constants.

   2. Express the electric field in terms of the applied voltage. For a conductor of length $L$, the uniform electric field inside is $E = V/L$. So:

   $v_d = \frac{e (V/L) \tau}{m} = \left(\frac{e\tau}{mL}\right) V$

   The quantity in parentheses is constant for a given conductor. So $v_d$ is directly proportional to $V$.

   3. Apply the change. If $V$ becomes $V/2$, then:

      $v_d' = \left(\frac{e\tau}{mL}\right) \cdot \frac{V}{2} = \frac{1}{2} \left(\frac{e\tau}{mL}\right) V = \frac{v_d}{2}$ …

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10. CBSE 2024Set FS1 markMCQ
    Q.If drift velocity of electron be $v_d$ and intensity of electric field E, then which relation among the following obeys Ohm's law?

    (i) $v_d \propto E^2$

    (ii) $v_d = \text{Constant}$

    (iii) $v_d \propto E$

    (iv) $v_d \propto \sqrt{E}$

    ›Reveal solutionSolution

    Drift velocity $v_d=\dfrac{eE\tau}{m}$ is directly proportional to the field $E$, which is exactly what Ohm's law requires — option (iii).

    Concept. Under an electric field $E$, free electrons acquire a steady drift velocity

    $v_d=\frac{eE\tau}{m},$

    where $\tau$ is the mean relaxation time.

    …

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11. CBSE 2024Set ANNUAL1 mark
    Q.Define the relaxation time of the free electrons drifting in a conductor.

    ›Reveal solutionSolution

    The average free time an electron travels between collisions with the lattice.

    …

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12. CBSE 2023Set 55/1/11 markMCQ
    Q.A steady current flows through a metallic wire whose area of cross-section $(A)$ increases continuously from one end of the wire to the other. The magnitude of drift velocity $(v_d)$ of the free electrons as a function of $A$ can be represented by :

    (a)

    (b)

    (c)

    (d)

    ›Reveal solutionSolution

    For a steady current, the product $A v_d$ is constant because $I = n e A v_d$ is fixed. Therefore $v_d \propto 1/A$, which is a rectangular hyperbola — option (a).

    

    The key to this question is understanding why drift velocity changes when the wire's cross-section changes — and that comes from the definition of steady current itself.

    When we say "a steady current flows", we mean that the same amount of charge passes through every cross-section of the wire per second. The wire is in series with itself: whatever charge flows past the thin end must also flow past the thick end in the same time. If it didn't, charge would pile up somewhere — and that would violate the steady-state condition.

    Now, current $I$ is given by:

    $I = n e A v_d$

    where $n$ is the free electron density (number per unit volume), $e$ is the electron charge, $A$ is the cross-sectional area at that point, and $v_d$ is the drift velocity.

    For a metallic wire, $n$ and $e$ are constants (same material throughout). And for a steady current, $I$ is constant along the wire. So:

    $n e A v_d = \text{constant}$

    which means:

    $A v_d = \text{constant}$

    Therefore:

    $v_d \propto \frac{1}{A}$

    This is the relationship we need.

    1. Identify the mathematical form. $v_d \propto 1/A$ is an inverse proportion. Its graph is a rectangular hyperbola — a curve that falls steeply when $A$ is small and flattens out as $A$ grows large. It never touches either axis (asymptotic behaviour).

    2. Check the options against this.

       • Option (a) shows exactly this: a curve that drops as $A$ increases, shaped like a hyperbola.
       • Option (b) is a straight line — that would mean $v_d \propto -A + \text{constant}$, which is wrong.
       • Option (c) is a horizontal line — that would mean $v_d$ is independent of $A$, which contradicts $A v_d = \text{constant}$.
       • Option (d) is a straight line through the origin — that would mean $v_d \propto A$, which is the opposite of what we have. …

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